Find out the oxidation number of chlorine in the following compounds:

1) NaClO₄ = +7

2) NaClO₃ = +5

3) NaClO = +1

4) KClO₂ = +3

5) Cl₂O₇ = +7

6) ClO₃ = +5

7) Cl₂O = +1

8) NaCl = -1

9) Cl₂ = 0

10) ClO₂ = +4

Determining the oxidation number of chlorine in various compounds involves understanding the rules of oxidation states and how they apply to different elements in a compound. Let's break down the oxidation states for each of the compounds you've listed, explaining the reasoning behind each value.

Oxidation States of Chlorine Compounds

The oxidation state of an element in a compound reflects its degree of oxidation or reduction. Here’s how we can find the oxidation number of chlorine in each of the compounds you mentioned:

1. Sodium Perchlorate (NaClO₄)

In NaClO₄, sodium (Na) has an oxidation state of +1, and oxygen (O) typically has an oxidation state of -2. There are four oxygen atoms, contributing a total of -8. To balance the charges in the compound:

• Let the oxidation state of Cl be x.
• +1 (Na) + x (Cl) + (-8) (from O) = 0.

Solving for x gives us:

x = +7. Thus, the oxidation number of chlorine in NaClO₄ is +7.

2. Sodium Chlorate (NaClO₃)

In NaClO₃, sodium is still +1, and there are three oxygen atoms contributing -6. Setting up the equation:

• +1 (Na) + x (Cl) + (-6) (from O) = 0.

Solving for x gives:

x = +5. Therefore, the oxidation number of chlorine in NaClO₃ is +5.

3. Sodium Hypochlorite (NaClO)

For NaClO, sodium remains +1, and there is one oxygen contributing -2. The equation is:

• +1 (Na) + x (Cl) + (-2) (from O) = 0.

Solving for x results in:

x = +1. Hence, the oxidation number of chlorine in NaClO is +1.

4. Potassium Chlorite (KClO₂)

In KClO₂, potassium has an oxidation state of +1, and there are two oxygen atoms contributing -4. The equation is:

• +1 (K) + x (Cl) + (-4) (from O) = 0.

Solving for x gives:

x = +3. Thus, the oxidation number of chlorine in KClO₂ is +3.

5. Dichlorine Heptoxide (Cl₂O₇)

In Cl₂O₇, there are two chlorine atoms and seven oxygen atoms contributing -14. The equation is:

• 2x (for Cl) + (-14) (from O) = 0.

Solving for x gives:

x = +7. Therefore, the oxidation number of chlorine in Cl₂O₇ is +7.

6. Chlorine Trioxide (ClO₃)

In ClO₃, there are three oxygen atoms contributing -6. The equation is:

• x (Cl) + (-6) (from O) = 0.

Solving for x results in:

x = +5. Thus, the oxidation number of chlorine in ClO₃ is +5.

7. Dichlorine Monoxide (Cl₂O)

In Cl₂O, there are two chlorine atoms and one oxygen contributing -2. The equation is:

• 2x (for Cl) + (-2) (from O) = 0.

Solving for x gives:

x = +1. Hence, the oxidation number of chlorine in Cl₂O is +1.

8. Sodium Chloride (NaCl)

In NaCl, sodium has an oxidation state of +1, and chlorine must balance this with -1. Thus, the oxidation number of chlorine in NaCl is -1.

9. Chlorine Gas (Cl₂)

In Cl₂, both chlorine atoms are in their elemental form, which means their oxidation state is 0.

10. Chlorine Dioxide (ClO₂)

In ClO₂, there are two oxygen atoms contributing -4. The equation is:

• x (Cl) + (-4) (from O) = 0.

Solving for x gives:

x = +4. Therefore, the oxidation number of chlorine in ClO₂ is +4.

By applying these rules and calculations, we can systematically determine the oxidation states of chlorine in various compounds. This understanding is crucial in fields like chemistry and environmental science, where redox reactions play a significant role.